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Question Number 117945 by bemath last updated on 14/Oct/20

f(x) = ∫ ((5x^8 +7x^6 )/((2x^7 +x^2 +1)^2 )) dx and f(0) = 0 , then f(1) = _

$$\mathrm{f}\left(\mathrm{x}\right)\:=\:\int\:\frac{\mathrm{5x}^{\mathrm{8}} +\mathrm{7x}^{\mathrm{6}} }{\left(\mathrm{2x}^{\mathrm{7}} +\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\:\mathrm{dx}\:\mathrm{and}\:\: \\ $$$$\mathrm{f}\left(\mathrm{0}\right)\:=\:\mathrm{0}\:,\:\mathrm{then}\:\mathrm{f}\left(\mathrm{1}\right)\:=\:\_\: \\ $$

Commented by bemath last updated on 14/Oct/20

thank you all sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{all}\:\mathrm{sir} \\ $$

Answered by john santu last updated on 14/Oct/20

f(x)=∫ ((5x^8 +7x^6 )/(x^(14) (2+x^(−5) +x^(−7) )^2 )) dx f(x)=∫ ((5x^(−6) +7x^(−8) )/((2+x^(−5) +x^(−7) )^2 )) dx letting φ = 2+x^(−5) +x^(−7) then dφ = −(5x^(−6) +7x^(−8) ) dx f(x) = ∫ ((−dφ)/φ^2 ) = −∫φ^(−2) dφ = (1/φ)+c f(x)= (1/(2+x^(−5) +x^(−7) )) + c f(x)= (x^7 /(2x^7 +x^2 +1)) + c since f(0) = 0 we get c = 0 thus f(1)= (1/4)

$${f}\left({x}\right)=\int\:\frac{\mathrm{5}{x}^{\mathrm{8}} +\mathrm{7}{x}^{\mathrm{6}} }{{x}^{\mathrm{14}} \left(\mathrm{2}+{x}^{−\mathrm{5}} +{x}^{−\mathrm{7}} \right)^{\mathrm{2}} }\:{dx}\: \\ $$$${f}\left({x}\right)=\int\:\frac{\mathrm{5}{x}^{−\mathrm{6}} +\mathrm{7}{x}^{−\mathrm{8}} }{\left(\mathrm{2}+{x}^{−\mathrm{5}} +{x}^{−\mathrm{7}} \right)^{\mathrm{2}} }\:{dx}\: \\ $$$${letting}\:\phi\:=\:\mathrm{2}+{x}^{−\mathrm{5}} +{x}^{−\mathrm{7}} \:{then}\: \\ $$$${d}\phi\:=\:−\left(\mathrm{5}{x}^{−\mathrm{6}} +\mathrm{7}{x}^{−\mathrm{8}} \right)\:{dx} \\ $$$${f}\left({x}\right)\:=\:\int\:\frac{−{d}\phi}{\phi^{\mathrm{2}} }\:=\:−\int\phi^{−\mathrm{2}} \:{d}\phi\:=\:\frac{\mathrm{1}}{\phi}+{c} \\ $$$${f}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{2}+{x}^{−\mathrm{5}} +{x}^{−\mathrm{7}} }\:+\:{c}\: \\ $$$${f}\left({x}\right)=\:\frac{{x}^{\mathrm{7}} }{\mathrm{2}{x}^{\mathrm{7}} +{x}^{\mathrm{2}} +\mathrm{1}}\:+\:{c}\: \\ $$$${since}\:{f}\left(\mathrm{0}\right)\:=\:\mathrm{0}\:{we}\:{get}\:{c}\:=\:\mathrm{0} \\ $$$${thus}\:{f}\left(\mathrm{1}\right)=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$

Answered by Dwaipayan Shikari last updated on 14/Oct/20

∫((5x^8 +7x^6 )/((2x^7 +x^2 +1)^2 ))dx =∫(((5/x^6 )+(7/x^8 ))/((2+(1/x^5 )+(1/x^7 ))^2 ))dx =−∫(dt/t^2 ) =(1/t)+C=(1/((2+(1/x^5 )+(1/x^7 )))) f(x)=(x^7 /(2x^7 +x^2 +1))+C⇒f(0)=0+C=0 f(1)=(1/(2.1+1+1))=(1/4)

$$\int\frac{\mathrm{5}{x}^{\mathrm{8}} +\mathrm{7}{x}^{\mathrm{6}} }{\left(\mathrm{2}{x}^{\mathrm{7}} +{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$=\int\frac{\frac{\mathrm{5}}{{x}^{\mathrm{6}} }+\frac{\mathrm{7}}{{x}^{\mathrm{8}} }}{\left(\mathrm{2}+\frac{\mathrm{1}}{{x}^{\mathrm{5}} }+\frac{\mathrm{1}}{{x}^{\mathrm{7}} }\right)^{\mathrm{2}} }{dx} \\ $$$$=−\int\frac{{dt}}{{t}^{\mathrm{2}} }\:\:=\frac{\mathrm{1}}{{t}}+{C}=\frac{\mathrm{1}}{\left(\mathrm{2}+\frac{\mathrm{1}}{{x}^{\mathrm{5}} }+\frac{\mathrm{1}}{{x}^{\mathrm{7}} }\right)} \\ $$$${f}\left({x}\right)=\frac{{x}^{\mathrm{7}} }{\mathrm{2}{x}^{\mathrm{7}} +{x}^{\mathrm{2}} +\mathrm{1}}+\mathrm{C}\Rightarrow{f}\left(\mathrm{0}\right)=\mathrm{0}+{C}=\mathrm{0} \\ $$$${f}\left(\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}.\mathrm{1}+\mathrm{1}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$

Answered by 1549442205PVT last updated on 14/Oct/20

Put x=(1/t)⇒dx=−(dt/t^2 ) f(x)=−∫(((5/t^8 )+(7/t^6 ))/(((2/t^7 )+(1/t^2 )+1)^2 )).(dt/t^2 ) =−∫((t^4 (5+7t^2 )dt)/((t^7 +t^5 +2)^2 ))=−∫((5t^4 +7t^6 )/((t^7 +t^5 +2)^2 ))dt Put t^7 +t^5 +2=u⇒du=(7t^6 +5t^4 )dt ⇒f(x(u))=−∫(du/u^2 )=(1/u)=(1/(t^7 +t^5 +2))+C ⇒f(x)=(1/((1/x^7 )+(1/x^5 )+2))=(x^7 /(2x^7 +x^2 +1))+C f(0)=0⇒C=0.Hence f(x)=(x^7 /(2x^7 +x^2 +1))⇒f(1)=(1/4)

$$\mathrm{Put}\:\mathrm{x}=\frac{\mathrm{1}}{\mathrm{t}}\Rightarrow\mathrm{dx}=−\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} } \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=−\int\frac{\frac{\mathrm{5}}{\mathrm{t}^{\mathrm{8}} }+\frac{\mathrm{7}}{\mathrm{t}^{\mathrm{6}} }}{\left(\frac{\mathrm{2}}{\mathrm{t}^{\mathrm{7}} }+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }+\mathrm{1}\right)^{\mathrm{2}} }.\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} } \\ $$$$=−\int\frac{\mathrm{t}^{\mathrm{4}} \left(\mathrm{5}+\mathrm{7t}^{\mathrm{2}} \right)\mathrm{dt}}{\left(\mathrm{t}^{\mathrm{7}} +\mathrm{t}^{\mathrm{5}} +\mathrm{2}\right)^{\mathrm{2}} }=−\int\frac{\mathrm{5t}^{\mathrm{4}} +\mathrm{7t}^{\mathrm{6}} }{\left(\mathrm{t}^{\mathrm{7}} +\mathrm{t}^{\mathrm{5}} +\mathrm{2}\right)^{\mathrm{2}} }\mathrm{dt} \\ $$$$\mathrm{Put}\:\mathrm{t}^{\mathrm{7}} +\mathrm{t}^{\mathrm{5}} +\mathrm{2}=\mathrm{u}\Rightarrow\mathrm{du}=\left(\mathrm{7t}^{\mathrm{6}} +\mathrm{5t}^{\mathrm{4}} \right)\mathrm{dt} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{x}\left(\mathrm{u}\right)\right)=−\int\frac{\mathrm{du}}{\mathrm{u}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{u}}=\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{7}} +\mathrm{t}^{\mathrm{5}} +\mathrm{2}}+\mathrm{C} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{7}} }+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{5}} }+\mathrm{2}}=\frac{\mathrm{x}^{\mathrm{7}} }{\mathrm{2x}^{\mathrm{7}} +\mathrm{x}^{\mathrm{2}} +\mathrm{1}}+\mathrm{C} \\ $$$$\mathrm{f}\left(\mathrm{0}\right)=\mathrm{0}\Rightarrow\mathrm{C}=\mathrm{0}.\mathrm{Hence} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}^{\mathrm{7}} }{\mathrm{2x}^{\mathrm{7}} +\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\Rightarrow\mathrm{f}\left(\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{4}} \\ $$

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